May 20th, 2017, 06:49 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Problems on Series
1. The largest Interval $I$ such that the series $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}} $ converges whenever $\displaystyle x \in I $ is equal to A)$ [1,1]$ B)$ [1,1)$ C)$ (1,1]$ D)$(1,1)$ My answer : Options A or B 2. Let $\displaystyle \sum a_n $ be a convergent series. Let $b_n=a_{n+1}  a_n$ for all $\displaystyle n \in N $. Then A) $\displaystyle \sum b_n $ should also be convergent and $\displaystyle (b_n) \rightarrow 0 $ as $\displaystyle n \rightarrow \infty. $ B) $\displaystyle \sum b_n $ need not be convergent but $\displaystyle (b_n) \rightarrow 0 $ as $\displaystyle n \rightarrow \infty. $ C) $\displaystyle \sum b_n $ is convergent but $(b_n)$ need not tend to zero as $\displaystyle n \rightarrow \infty. $ D) None of the above statements is true. My answer : Option A 3. Which of the following series converge ? A) $\displaystyle \sum_{n=1}^{\infty} (\frac{\log n}{n^{1+2\epsilon}} $ B) $\displaystyle \sum_{n=1}^{\infty} (\frac{(\log n)^2}{n^{1+2\epsilon}} $ C) $\displaystyle \sum_{n=1}^{\infty} (\frac{n^2+1}{n^3+n}) $ D) $\displaystyle \sum_{n=1}^{\infty} (1+\frac{1}{n})^n $ I checked Option C, it is convergent and Option D is not convergent. Kindly check if Option A or B are Convergent ? Help me to solve series with logarithms. Kindly check and let me know If I'm wrong. Thanks Last edited by skipjack; May 20th, 2017 at 09:43 PM. 
May 20th, 2017, 10:25 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,379 Thanks: 2011 
1. (B) 2. (A) 3. The parentheses don't balance in (A) and (B), and ϵ is undefined. (C) and (D) are divergent. 
May 20th, 2017, 10:41 PM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
Sorry to confuse you. Can you please check the below series now ? A) $\displaystyle \sum_{n=1}^{\infty} (\frac{\log n}{n^{1+2e}}) $ B) $\displaystyle \sum_{n=1}^{\infty} (\frac{(\log n)^2}{n^{1+2e}}) $  
May 20th, 2017, 10:56 PM  #4  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
It's confusing for me to estimate a series converges or not. Thanks  
May 21st, 2017, 08:44 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra 
C is precisely $\sum \frac1n$ and thus divergent. The terms of D tend to $e$ and not $0$, so that is divergent. 

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